Three Dimensional Geometry Question 123
Question: The equation of the plane passing through the lines $ \frac{x-4}{1}=\frac{y-3}{1}=\frac{z-2}{2} $ and $ \frac{x-3}{1}=\frac{y-2}{-4}=\frac{z}{5} $ is
Options:
A) $ 11x-y-3z=35 $
B) $ 11x+y-3z=35 $
C) $ 11x-y+3z=35 $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ a,(x-4)+b,(y-3)+c,(z-2)=0 $
$ \therefore ,a+b+2c=0 $ and $ a-4b+5c=0 $ $ \frac{a}{5+8}=\frac{b}{2-5}=\frac{c}{-4-1}=k $ $ \frac{a}{13}=\frac{b}{-3}=\frac{c}{-5}=k $ Therefore, the required equation of plane is $ -13x-3y-5z+33=0 $ .
BETA
